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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p><dfn class="terminology">Proof</dfn> Consider</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq4_6.html">
\begin{equation}
k_1 e^{r_1 x}+k_2 e^{r_2 x}+\cdots+k_n e^{r_n x}=0.\tag{4.2.1}
\end{equation}
</div>
<p class="continuation">We multiply <span class="process-math">\(e^{-r_1 x}\)</span> on both sides of the equation, we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq4_6.html">
\begin{equation*}
k_1+k_2 e^{(r_2-r_1) x}+k_3 e^{(r_3-r_1) x}+\cdots+k_n e^{(r_n-r_1) x}=0.
\end{equation*}
</div>
<p class="continuation">Differentiation gives</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq4_6.html">
\begin{equation*}
k_2 (r_2-r_1) e^{(r_2-r_1) x}+k_3 (r_3-r_1) e^{(r_3-r_1) x}+\cdots+k_n (r_n-r_1) e^{(r_n-r_1) x}=0.
\end{equation*}
</div>
<p class="continuation">Multiplying <span class="process-math">\(e^{-(r_2-r_1) x}\)</span> on both sides,</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq4_6.html">
\begin{equation*}
k_2 (r_2-r_1)+k_3 (r_3-r_1) e^{(r_3-r_2) x}+\cdots+k_n (r_n-r_1) e^{(r_n-r_2) x}=0.
\end{equation*}
</div>
<p class="continuation">Differentiation gives</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq4_6.html">
\begin{equation*}
k_3 (r_3-r_1)(r_3-r_2) e^{(r_3-r_2) x}+\cdots+k_n (r_n-r_1) (r_n-r_2) e^{(r_n-r_2) x}=0.
\end{equation*}
</div>
<p class="continuation">We do this continuously and finally have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq4_6.html">
\begin{equation*}
k_n (r_n-r_1)(r_n-r_2)\cdots(r_n-r_{n-1}) e^{(r_n-r_{n-1}) x}=0.
\end{equation*}
</div>
<p class="continuation">The last equation gives <span class="process-math">\(k_n=0\text{.}\)</span> Then (<a href="" class="xref" data-knowl="./knowl/eq4_6.html" title="Equation 4.2.1">(4.2.1)</a>) becomes</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq4_6.html">
\begin{equation*}
k_1 e^{r_1 x}+k_2 e^{r_2 x}+\cdots+k_n e^{r_{n} x}=0.
\end{equation*}
</div>
<p class="continuation">By the same procedure, it can be shown that</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq4_6.html">
\begin{equation*}
k_{n-1}=0,\quad k_{n-2}=0,\quad k_1=0.
\end{equation*}
</div>
<p class="continuation">This is to say <span class="process-math">\(e^{r_1 x}, e^{r_2 x},\cdots, e^{r_n x}\)</span> are linear independent. The general solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq4_6.html">
\begin{equation*}
y=C_1 e^{r_1 x}+C_2 e^{r_2 x}+\cdots C_n e^{r_n x}.
\end{equation*}
</div>
<span class="incontext"><a href="sec4_2.html#p-157" class="internal">in-context</a></span>
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